Advertisements
Advertisements
प्रश्न
Evaluate the following limits:
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`
Advertisements
उत्तर
We know `lim_(x -> 0) (sin x)/x` = 1
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = lim_(alpha -> 0) (sin (alpha^"n"))/(1/alpha^"n" (alpha^"n")) xx (alpha^(1/"m") * alpha^"m")/(sin alpha)^"m"`
= `lim_(alpha -> 0) alpha^"n" * (sin(alpha^"n"))/alpha^"n" xx 1/alpha^"m" * 1/(((sinalpha)^"m")/(alpha^"m"))`
= `lim_(alpha -> 0) alpha^("n" - "m") *(sin(alpha^"n"))/alpha^"n" xx 1/((sinalpha)/alpha)^"m"`
= `(lim_(alpha -> 0) alpha^("n" - "m")) xx (lim_(alpha^"n" -> 0) (sin(alpha^"n"))/alpha^"n") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
Case (i) m = n
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("m" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/(lim_(alpha -> 0) ((sin alpha)/alpha)^"m")`
= `(lim_(alpha -> 0) alpha^0) xx 1 xx 1/1`
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = 1 xx 1 xx 1` = 1
Case (ii) m > n then n – m < 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(apha^"m" -> 0) (sin(alpha^"m"))/alpha^"m") xx 1/((lim_(alpha-> 0) ((sin alpha)/alpha)^"m"))`
= `lim_(alpha -> 0) 1/(alpha^("m" - "n")) xx 1 xx 1/1`
= `oo xx 1 xx 1 = oo`
Since `lim_(alpha -> 0) 1/(alpha^("m" - "n")) = lim_(alpha -> 0) (1/0)^("m" - "n") = oo`
Case (iii) m < n then n – m > 0
`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = (lim_(alpha -> 0) alpha^("n" - "m")) (lim_(alpha^"m" -> 0) (sin(alpha^"m"))/(alpha^"m")) xx 1/(lim_(alpha-> 0) ((sin alpha)/alpha)^"m"`
= `(0)^("n" - "m") xx 1 xx 1`
= 0
∴ `lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m" = {{:(1, "if", "m" = "n"),(oo, "if", "m" > "n"),(0, "if", m < n):}`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit:
`lim_(y -> -3) [(y^5 + 243)/(y^3 + 27)]`
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limit :
`lim_(x -> 7) [(x^3 - 343)/(sqrt(x) - sqrt(7))]`
In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.
`lim_(x -> 1) (x^2 + x + 1)` = 3
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 1) f(x)` where `f(x) = {{:(x^2 + 2",", x ≠ 1),(1",", x = 1):}`
In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?
`lim_(x -> 3) 1/(x - 3)`
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Evaluate : `lim_(x -> 3) (x^2 - 9)/(x - 3)` if it exists by finding `f(3^-)` and `f(3^+)`
Evaluate the following limits:
`lim_(x -> 2) (1/x - 1/2)/(x - 2)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(x^2 + 1) - 1)/(sqrt(x^2 + 16) - 4)`
Evaluate the following limits:
`lim_(x ->oo) (x^3/(2x^2 - 1) - x^2/(2x + 1))`
Evaluate the following limits:
`lim_(x -> 0) (sinalphax)/(sinbetax)`
Evaluate the following limits:
`lim_(x -> 0) (tan 2x)/x`
Evaluate the following limits:
`lim_(x -> 0) (1 - cos^2x)/(x sin2x)`
Evaluate the following limits:
`lim_(x -> 0) (sqrt(2) - sqrt(1 + cosx))/(sin^2x)`
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
Choose the correct alternative:
`lim_(x - pi/2) (2x - pi)/cos x`
Choose the correct alternative:
`lim_(x -> 0) sqrt(1 - cos 2x)/x`
Choose the correct alternative:
`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =
`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.
